Does $\sum _{n=2}^{\infty}\frac{1}{\ln\left(n!\right)}$ converge?

Question

I spend too much time on this one,and I just can't find a way to prove anything. Please, will someone help me?

Does $$ \sum _{n=2}^{\infty}\frac{1}{\ln\left(n!\right)}$$ converge?

P.S. Does anybody knows where can I find (online) these kind of tasks (I mean, with this kind of difficulty). Thank you (in advance). :)

Well, you don't want to start at n=1 I would assume – imranfat Aug 23 '16 at 21:07 — imranfat, Aug 23 '16 at 21:07

score 7 · Accepted Answer · edited Apr 13 '17 at 12:21

7

No, it does not.

It is standard to see that $\ln n! \sim_{n\to\infty} n\ln n$, so that the series with positive terms $\sum_n \frac{1}{\ln n!}$ and $\sum_n \frac{1}{n \ln n}$ have same nature by theorems of comparison.

But the latter diverges, as a Bertrand series.$^{(\dagger)}$

Theorem. (Bertrand series) The series $\sum_{n=2}^\infty \frac{1}{n^a(\ln n)^b}$ converges if, and only if, (i) $a>1$ or (ii) $a=1$ and $b>1$.

edited Apr 13 '17 at 12:21

Community

1

answered Aug 23 '16 at 21:09

Clement C.

67,323

Standard with or without Stirling ? – Gabriel Romon Aug 23 '16 at 21:10
1

Might want to reference Stirling's approximation . – Alexis Olson Aug 23 '16 at 21:10
1

@AlexisOlson See the link provided. No need for Stirling, actually. – Clement C. Aug 23 '16 at 21:12
@LeGrandDODOM without: Look at $(\sum_{k=1}^n \ln k)/(n\ln n)$ with Stolz-Cesaro. – zhw. Aug 23 '16 at 21:14
@zhw the integral method he referenced is even more elementary – Gabriel Romon Aug 23 '16 at 21:16
Without invoking Bertrand series, one can also show that $\sum_{n=2}^N \frac{1}{n\ln n}$ diverges by (yet another) integral comparison, since $\int_2^N \frac{dx}{x\ln x} = \ln\ln N - \ln\ln 2$. – Clement C. Aug 23 '16 at 21:20
@LeGrandDODOM Disagree, SC is simpler (if you know it) (IMO of course) – zhw. Aug 23 '16 at 21:25
2

@zhw. The comments are not really the place for a debate on "what is simpler, and what does 'simple' mean in terms of a tradeoff between conciseness of the proof and level of the tools used in it." As for the question itself, all that is needed is that $\ln n! \leq n\ln n$, which is actually immediate since $\ln n! = \sum_{k=1}^n \ln k \leq \sum_{k=1}^n \ln n$. – Clement C. Aug 23 '16 at 21:28
@ClementC. Actually, zhw was responding to a question posted by LeGrandDODOM. It was the latter who began the debate of which method was simpler. ;-)) – Mark Viola Aug 23 '16 at 21:56
Fait enough :) @Dr.MV – Clement C. Aug 23 '16 at 21:57
And +1 for the solution – Mark Viola Aug 23 '16 at 22:09

score 4 · Answer 2 · answered Aug 23 '16 at 22:11

For all $n >1$, we have $n^n > n!$. Therefore, $\displaystyle \frac{1}{n^n} < \frac{1}{n!}$, and further $\displaystyle \frac{1}{\ln(n^n)} < \frac{1}{\ln(n!)}$ since $\ln(x)$ is an increasing function. By a property of logarithms, we have $\displaystyle \frac{1}{\ln(n^n)} = \frac{1}{n\ln(n)}$.

Since $\displaystyle \sum_{n=2}^\infty \frac{1}{n \ln(n)}$ can be shown to diverge with, e.g., the integral test, it follows that $\displaystyle \sum_{n=2}^\infty \frac{1}{\ln(n!)}$ diverges per the comparison test.

Does $\sum _{n=2}^{\infty}\frac{1}{\ln\left(n!\right)}$ converge?

2 Answers2